What is the pressure in millimeters of mercury of 0.0115mol of helium gas with a volume of 206mL at a temperature of 25∘C?
To find the pressure in millimeters of mercury (mmHg) of a gas, you can use the Ideal Gas Law equation:
PV = nRT
Where:
P is the pressure (in units of Pascal or Pa, but we can convert it to mmHg later)
V is the volume of the gas (in units of liters)
n is the number of moles of the gas
R is the ideal gas constant (8.314 J/(mol·K) or 0.0821 L·atm/(mol·K))
T is the temperature (in units of Kelvin or K)
To get the answer, we need to convert the given values to the appropriate units and then solve for P. Let's go through the steps.
1. Convert the volume from milliliters (mL) to liters (L):
206 mL = 206 / 1000 L = 0.206 L
2. Convert the temperature from Celsius (°C) to Kelvin (K):
T(°C) + 273.15 = T(K)
25°C + 273.15 = 298.15 K
3. Substitute the known values into the equation:
PV = nRT
P * 0.206 L = 0.0115 mol * 0.0821 L·atm/(mol·K) * 298.15 K
4. Solve for P:
P = (0.0115 mol * 0.0821 L·atm/(mol·K) * 298.15 K) / 0.206 L
P ≈ 4.03 atm
5. Convert the pressure from atm to mmHg:
1 atm = 760 mmHg
P (mmHg) = 4.03 atm * 760 mmHg/atm
P ≈ 3065 mmHg
Therefore, the pressure in millimeters of mercury of 0.0115 mol of helium gas with a volume of 206 mL at a temperature of 25°C is approximately 3065 mmHg.